Решите методом замены (x^2+x-2)(x^2+x)=24 и задание с параметром при каких значениях “ a”x1*x2=3 x^2+(a+3)x+a^2-2a=0

x=-3, x=2

Объяснение:

Замена:

Обратная замена:

x\in \varnothing\\\\2)\;t=6,\; x^2+x=6\\x^2+x-6=0\\x^2-2x+3x-6=0\\x(x-2)+3(x-2)=0\\(x-2)(x+3)=0\\\left[\begin{array}{c}x=2\\x=-3\end{array}\right;" class="latex-formula" id="TexFormula4" src="https://tex.z-dn.net/?f=1%29%5C%3Bt%3D-4%2C%5C%3B%20x%5E2%2Bx%3D-4%5C%5Cx%5E2%2Bx%2B4%3D0%5C%5CD%3D1-16%3C0%5C%5C%3D%3Ex%5Cin%20%5Cvarnothing%5C%5C%5C%5C2%29%5C%3Bt%3D6%2C%5C%3B%20x%5E2%2Bx%3D6%5C%5Cx%5E2%2Bx-6%3D0%5C%5Cx%5E2-2x%2B3x-6%3D0%5C%5Cx%28x-2%29%2B3%28x-2%29%3D0%5C%5C%28x-2%29%28x%2B3%29%3D0%5C%5C%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx%3D2%5C%5Cx%3D-3%5Cend%7Barray%7D%5Cright%3B" title="1)\;t=-4,\; x^2+x=-4\\x^2+x+4=0\\D=1-16<0\\=>x\in \varnothing\\\\2)\;t=6,\; x^2+x=6\\x^2+x-6=0\\x^2-2x+3x-6=0\\x(x-2)+3(x-2)=0\\(x-2)(x+3)=0\\\left[\begin{array}{c}x=2\\x=-3\end{array}\right;">

Уравнение решено!